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Math Help - is it correct?

  1. #1
    kansai
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    need help gr12

    can any one tell me how to do question 1, 3,4,5,6,7,8? and when u have answer plz remember to explain ur steps thnx

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  2. #2
    Super Member malaygoel's Avatar
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    hello, I could give answer to your first question
    when a square is inscribed in a circle, the length of the diagnol of the square =the diameter of the circle
    2^(1/2)side of square = diagonal of square
    area of square = 1/2 * diagonal * diagonal
    S1 = 2 (diagonal of S1 =2)
    when a circle is inscribed in a square, length of square = diameter of circle
    diameter of O2= length of S1 = 2^(-1/2)* diagonal of S1 = 2^(1/2)
    S2 = 1/2 * 2 = 1
    Sn = 2*[(1/2)^(n-1)]
    Last edited by malaygoel; June 5th 2006 at 06:03 PM. Reason: A silly mistake
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  3. #3
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    Quote Originally Posted by kansai
    can any one do 3,4,5,6,7,8??????
    2)
    You know that,
    t_1=-2
    t_2=1
    And that,
    t_3=2t_2-t_1
    thus,
    t_3=2(1)-(-2)=2+2=4
    And that,
    t_4=2t_3-t_2
    thus,
    t_4=2(4)-1=8-1=7
    And that,
    t_5=2t_4-t_3
    thus,
    t_5=2(7)-1=13
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  4. #4
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    Hello, kansai!

    Here's #5 . . .


    5. For the matrices: A = \begin{bmatrix}1&5&0\\3&0&4\\2&0&-1\end{bmatrix} and B = \begin{bmatrix}2&0&-1\\1&0&1\\3&-2&0\end{bmatrix}

    (a) State the value of: \text{(i) }a_{12}\qquad\text{(ii) }a_{22}\qquad\text{(iii) }a_{31}

    (b) Calculate: \text{(i) }A - 3B\qquad\text{(ii) }2A + B\qquad\text{(iii) }A'

    \text{(a) }\;a_{12} means the element in the 1^{st} row and the 2^{nd}column. .Hence: a_{12} = 5
    . . . .Similarly: a_{22} = 0,\;a_{31} = 2

    The elements of matrix B would have been written: b_{12},\;b_{22},\;b_{31}



    \text{(b)(i) }A - 3B \:=\:\begin{bmatrix}1&5&0\\3&0&4\\2&0&-1\end{bmatrix} - 3\cdot\begin{bmatrix}2&0&-1\\1&0&1\\3&-2&0\end{bmatrix}

    . . . = \;\begin{bmatrix}1&5&0\\3&0&4\\2&0&-1\end{bmatrix} - \begin{bmatrix}6&0&-3\\3&0&3\\9&-6&0\end{bmatrix} \;=\;\begin{bmatrix}-5&5&3\\0&0&1\\-7&6&-1\end{bmatrix}



    \text{(b)(ii) }2A + B \;= \;2\begin{bmatrix}1&5&0\\3&0&4\\2&0&-1\end{bmatrix} + \begin{bmatrix}2&0&-1\\1&0&1\\3&-2&0\end{bmatrix}

    . . . =\;\begin{bmatrix}2&10&0\\6&0&8\\4&0&-2\end{bmatrix} + \begin{bmatrix}2&0&-1\\1&0&1\\3&-2&0\end{bmatrix} \;= \;\begin{bmatrix}4&10&-1\\7&0&9\\7&-2&-2\end{bmatrix}



    \text{(b)(iii) }A' . . . I assume this means the inverse of A.

    The steps are long and messy . . . so I'll omit them.

    . . . A'\;=\;\frac{1}{55}\begin{bmatrix}0&5&20\\11&-1&-4\\0&10&-15\end{bmatrix}

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  5. #5
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    Hello, kansai!

    Here's #3 . . .

    For the map given on the right
    (a) Represent the map with a network.
    (b) Find the degree of each vertex.
    (c) State whether the network is traceable and explain the reason.
    Code:
    (a)
    
            A - - - - - B
            |         / | \
            |       /   |  \
            |     C     |   \
            |   /       |    \
            | /         |     \
            D - - - - - E - - -F
    (b) A: degree 2
    . . .B: degree 4
    . . .C: degree 2
    . . .D: degree 3
    . . .E: degree 3
    . . .F: degree 2


    (c) A network is traceable if it has at most two vertices of odd degree.

    This network has two odd vertices (D and E) and hence is traceable.
    Last edited by Soroban; June 5th 2006 at 04:53 AM.
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  6. #6
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    Hello, kansai!

    I don't "trust" #4 . . . It has some disturbing errors and the meaning is unclear.
    . . Can you provide the original wording?

    4. The NYS school has four committees.
    Each of these committees meet once a month.

    Membership on these committees is as follows:
    . . Committee A: Anna, John, Steven, Tom
    . . Committee B: Debra, Jane John
    . . Committee C: Annie, Debra, Larry, Tom
    . . Committee D: Bobby, Debra, Grace, Mary, Steven
    . . Committee E: Anna, Larry, Tom.

    (a) Draw a network to illustrate the connectivity among the committees. .?

    (b) Design a schedule with minimum timeslots for the committee meeting without conflicts. .??
    A petty point: I assume it refers to The "NYS school system".


    I would assume that Annie is the same person as Anna, that someone mistyped the names.
    . . But do I dare make that assumption?


    The most glaring errors is that there are five committees.


    (a) Connectivity is a term used in network theory, but not like this.
    . . I assume they mean "common memberships" ?


    (b) Minimum timeslots is a very sloppy term.

    Does it mean "the shortest meetings"?
    . . How about: "I call the meeting to order. .I will entertain a motion to adjourn . . ."

    Does it mean "the shortest time-span" for the four (five?) meetings?
    . .Schedule them consecutively with a one-minute break in between
    . . (to allow the ten people to switch seats).

    It is obvious that no two committees can meet at the same time.
    . . So what is the question?


    If I am correct about part (a), this should suffice:
    Code:
                      B
                    / | \
                  /   |   \
                /     |     \
              A - - - + - - - D
             /  \     |     /
            /     \   |   /
           /        \ | /
          E - - - - - C
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  7. #7
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    URGENT can any one do question 6,7,8?

    URGENT can any one do question 6,7,8?? and plz show ur steps and explation..............thnx
    Last edited by MathGuru; June 7th 2006 at 11:32 AM.
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  8. #8
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by malaygoel
    hello, I could give answer to your first question
    when a square is inscribed in a circle, the length of the diagnol of the square =the diameter of the circle
    area of square = 1/2 * diagonal * diagonal
    S1 = 1/2
    when a circle is inscribed in a square, length of square = diameter of circle
    diameter of O2= (1/2)^(1/2)
    S2 = 1/2 * 1/2 = 1/4
    Sn = (1/2)^n
    The questions to this test were already answered in General high school math-need help gr12
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  9. #9
    MHF Contributor Quick's Avatar
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    Question 1

    From my understanding of this question it works out like this.

    the diagnol of the square (d) = the diameter of the circle (d), so the area =

     s = the length of one of the square's side

     d=\sqrt{ s^2 +s^2}
     d^2=s^2 +s^2
     d^2=2s^2
     \frac{d^2}{2}=s^2 since s^2 is the area of the square this is the answer for  S_1 .
    Now the next circle has the diameter equal to the height of the square, so...
    \sqrt{\frac{d^2}{2}}=s=d_2
    and now you apply the first formula to find the area of  S_2
     \sqrt{\frac{d^2}{2}}=\sqrt{s^2 +s^2}
     \sqrt{\left(\frac{d^2}{2}\right)^2}=s^2 +s^2
     \frac{d^2}{2}=2s^2
     \frac{\frac{d^2}{2}}{2}=s^2
     \frac{d^2}{4}=s^2
     \frac{d^2}{4}=s^2

    This shows us that the area of  S_n = \frac{d^2}{2^n}

    Considering me and malaygoel got different answers, could a third party check my work?
    Last edited by Quick; June 5th 2006 at 06:27 PM.
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  10. #10
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Quick
    From my understanding of this question it works out like this.

    the diagnol of the square (d) = the diameter of the circle (d), so the area =

     s = the length of one of the square's side

     d=\sqrt{ s^2 +s^2}
     d=s +s
     d=2s
    how did you get d=s+s
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  11. #11
    MHF Contributor Quick's Avatar
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    how did you get d=s+s
    I made a mistake, which I corrected on the post, but I suppose it doesn't matter since he was asking for #6 instead of #1!
    Last edited by Quick; June 5th 2006 at 06:25 PM.
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  12. #12
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Quick

    Considering me and malaygoel got different answers, could a third party check my work?
    I have corrected my answer in the post. Could you please check it?
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  13. #13
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    diameter of O2= length of S1 = 2^(-1/2)* diagonal of S1 = 2^(1/2)
    it might be that I'm tired, buy I can't seem to think where you got the 2^(-1/2) from. However, you can see our equations differ,
    mine=  \frac{d^2}{2^n} yours=  2\left(\frac{1}{2^{(n-1)}}\right)

    However, there is a certain value of d that would make our equations the same.

     \frac{d^2}{2^n}=2\left(\frac{1}{2^{(n-1)}}\right)
     \frac{d^2}{2^n}=\frac{2}{2^{(n-1)}}
     \frac{2^2}{2^n}=\frac{2}{2^{(n-1)}}
     \frac{4}{2^n}=\frac{2}{2^{(n-1)}}
     \frac{2}{2^{(n-1)}}=\frac{2}{2^{(n-1)}}

    but why would you say the diameter is 2?
    Last edited by Quick; June 5th 2006 at 07:11 PM.
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  14. #14
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    friend got these answer for #1

    plz tell me is it correct?


    area of 1st square = 2
    area of 2nd square= 1
    area of 3rd square= 2^-1
    area of Sn square= 2^2-n
    is my friends answer correct?
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  15. #15
    MHF Contributor Quick's Avatar
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    assuming that the diameter of the first circle is 2, then yes.


     \frac{d^2}{2^n}
     \frac{2^2}{2^n}
     \frac{4}{2^n}
     \frac{4}{2^n}
     \frac{1}{2^{(n-2)}}
     2^{(-1)(n-2)}
     2^{(-n+2)}
     2^{(2-n)}
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